{"id":148,"date":"2023-07-25T20:37:39","date_gmt":"2023-07-25T20:37:39","guid":{"rendered":"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/"},"modified":"2023-07-25T20:37:39","modified_gmt":"2023-07-25T20:37:39","slug":"lewis-n2o4un-yapisi","status":"publish","type":"post","link":"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/","title":{"rendered":"5 ad\u0131mda n2o4 lewis yap\u0131s\u0131 (resimlerle)"},"content":{"rendered":"
\"Lewis<\/figure>\n

Yukar\u0131daki g\u00f6rseli zaten g\u00f6rd\u00fcn\u00fcz de\u011fil mi?<\/p>\n

Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m.<\/p>\n

N2O4’\u00fcn Lewis yap\u0131s\u0131nda, merkezde d\u00f6rt oksijen (O) atomu ile \u00e7evrelenmi\u015f iki nitrojen (N) atomu bulunur. \u0130ki oksijen atomu \u00e7ift ba\u011fl\u0131, di\u011fer iki oksijen atomu ise nitrojen atomuna tek ba\u011fl\u0131. Tek ba\u011fl\u0131 oksijen (O) atomlar\u0131nda 3 ortak \u00e7ift, \u00e7ift ba\u011fl\u0131 oksijen (O) atomlar\u0131nda ise 2 ortak \u00e7ift bulunur.<\/mark><\/em><\/strong><\/p>\n

N2O4’\u00fcn Lewis yap\u0131s\u0131n\u0131n yukar\u0131daki g\u00f6r\u00fcnt\u00fcs\u00fcnden hi\u00e7bir \u015fey anlamad\u0131ysan\u0131z, benimle kal\u0131n ve N2O4’\u00fcn<\/a> Lewis yap\u0131s\u0131n\u0131n \u00e7izilmesine ili\u015fkin ayr\u0131nt\u0131l\u0131 ad\u0131m ad\u0131m a\u00e7\u0131klamay\u0131 alacaks\u0131n\u0131z.<\/p>\n

O halde N2O4’\u00fcn Lewis yap\u0131s\u0131n\u0131<\/a> \u00e7izme ad\u0131mlar\u0131na ge\u00e7elim.<\/p>\n

N2O4’\u00fcn Lewis yap\u0131s\u0131n\u0131 \u00e7izme ad\u0131mlar\u0131<\/strong><\/h2>\n

Ad\u0131m 1: N2O4 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 bulun<\/strong><\/h3>\n

Bir N2O4 molek\u00fcl\u00fcndeki<\/a> toplam de\u011ferlik elektron<\/a> say\u0131s\u0131n\u0131 bulmak i\u00e7in \u00f6ncelikle oksijen atomunun yan\u0131 s\u0131ra nitrojen atomunda da bulunan de\u011ferlik elektronlar\u0131n\u0131 bilmeniz gerekir.
(De\u011ferlik elektronlar\u0131 herhangi bir atomun en d\u0131\u015f y\u00f6r\u00fcngesinde<\/a> bulunan elektronlard\u0131r.)<\/p>\n

Burada size periyodik tabloyu kullanarak nitrojenin ve oksijenin de\u011ferlik elektronlar\u0131n\u0131 nas\u0131l kolayca bulaca\u011f\u0131n\u0131z\u0131 anlataca\u011f\u0131m.<\/p>\n

N2O4 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong><\/p>\n

\u2192 Azot atomunun verdi\u011fi de\u011ferlik elektronlar\u0131:<\/strong> <\/p>\n

\"\"<\/figure>\n

Azot periyodik tablonun 15. grubunda yer alan bir elementtir. [1]<\/sup><\/a> Bu nedenle nitrojende bulunan de\u011ferlik elektronlar\u0131 5’tir<\/strong> . <\/p>\n

\"\"<\/figure>\n

Yukar\u0131daki resimde g\u00f6sterildi\u011fi gibi nitrojen atomunda bulunan 5 de\u011ferlik elektronunu g\u00f6rebilirsiniz.<\/p>\n

\u2192 Oksijen atomunun verdi\u011fi de\u011ferlik elektronlar\u0131:<\/strong> <\/p>\n

\"\"<\/figure>\n

Oksijen periyodik tablonun 16. grubunda yer alan bir elementtir. [2]<\/sup><\/a> Bu nedenle oksijende bulunan de\u011ferlik elektronlar\u0131 6’d\u0131r<\/strong> . <\/p>\n

\"\"<\/figure>\n

Yukar\u0131daki resimde g\u00f6sterildi\u011fi gibi oksijen atomunda bulunan 6 de\u011ferlik elektronunu g\u00f6rebilirsiniz.<\/p>\n

Bu y\u00fczden,<\/p>\n

N2O4 molek\u00fcl\u00fcndeki toplam de\u011ferlik elektronlar\u0131<\/strong> = 2 nitrojen atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 + 4 oksijen atomu taraf\u0131ndan ba\u011f\u0131\u015flanan de\u011ferlik elektronlar\u0131 = 5(2) + 6(4) = 34<\/strong> .<\/p>\n

Ad\u0131m 2: Merkez atomu se\u00e7in<\/strong><\/h3>\n

Merkez atomu se\u00e7mek i\u00e7in en az elektronegatif<\/a> atomun merkezde kald\u0131\u011f\u0131n\u0131 unutmamal\u0131y\u0131z.<\/p>\n

\u015eimdi burada verilen molek\u00fcl N2O4’t\u00fcr ve nitrojen atomlar\u0131 (N) ve oksijen4 atomu (O) i\u00e7erir. <\/p>\n

\"\"<\/figure>\n

Yukar\u0131daki periyodik tabloda nitrojen (N) atomu ve oksijen (O) atomunun elektronegatiflik de\u011ferlerini g\u00f6rebilirsiniz.<\/p>\n

Azot (N) ve oksijenin (O) elektronegatiflik de\u011ferlerini kar\u015f\u0131la\u015ft\u0131r\u0131rsak azot atomu daha az elektronegatiftir<\/a> .<\/p>\n

Burada nitrojen (N) atomlar\u0131 merkez atom, oksijen (O) atomlar\u0131 ise d\u0131\u015f atomlard\u0131r. <\/p>\n

\"N2O4<\/figure>\n

Ad\u0131m 3: Her atomu aralar\u0131na bir \u00e7ift elektron yerle\u015ftirerek ba\u011flay\u0131n<\/strong><\/h3>\n

\u015eimdi N2O4 molek\u00fcl\u00fcnde elektron \u00e7iftlerini<\/a> iki nitrojen (N) atomu aras\u0131na ve nitrojen (N) ile oksijen (O) atomlar\u0131 aras\u0131na yerle\u015ftirmeniz gerekiyor. <\/p>\n

\"N2O4<\/figure>\n

Bu, bir N2O4 molek\u00fcl\u00fcnde bu atomlar\u0131n birbirine kimyasal olarak ba\u011fland\u0131\u011f\u0131n\u0131<\/a> g\u00f6sterir.<\/p>\n

Ad\u0131m 4: D\u0131\u015f Atomlar\u0131 Kararl\u0131 Hale Getirin<\/strong><\/h3>\n

Bu ad\u0131mda d\u0131\u015f atomlar\u0131n kararl\u0131l\u0131\u011f\u0131n\u0131 kontrol etmeniz gerekir.<\/p>\n

Burada N2O4 molek\u00fcl\u00fcn\u00fcn \u00e7iziminde d\u0131\u015ftaki atomlar\u0131n oksijen atomlar\u0131 oldu\u011funu g\u00f6rebilirsiniz.<\/p>\n

Bu harici oksijen atomlar\u0131 bir oktet<\/a> olu\u015fturur ve bu nedenle stabildir. <\/p>\n

\"N2O4<\/figure>\n

Ek olarak, 1. ad\u0131mda N2O4 molek\u00fcl\u00fcnde bulunan toplam de\u011ferlik elektronu say\u0131s\u0131n\u0131 hesaplad\u0131k.<\/p>\n

N2O4 molek\u00fcl\u00fcn\u00fcn toplam 34 de\u011ferlik elektronu<\/strong> vard\u0131r ve bu de\u011ferlik elektronlar\u0131n\u0131n t\u00fcm\u00fc yukar\u0131daki N2O4 diyagram\u0131nda kullan\u0131lm\u0131\u015ft\u0131r.<\/p>\n

Bu nedenle merkez atomda tutulacak daha fazla elektron \u00e7ifti yoktur.<\/p>\n

\u015eimdi bir sonraki ad\u0131ma ge\u00e7elim.<\/p>\n

Ad\u0131m 5: Merkezi atomdaki sekizliyi kontrol edin. Okteti yoksa yaln\u0131z \u00e7ifti \u00e7ift ba\u011f veya \u00fc\u00e7l\u00fc ba\u011f olu\u015fturacak \u015fekilde hareket ettirin.<\/strong><\/h3>\n

Bu ad\u0131mda nitrojen (N) atomlar\u0131n\u0131n kararl\u0131 olup olmad\u0131\u011f\u0131n\u0131 kontrol etmeniz gerekir.<\/p>\n

Merkezi nitrojen (N) atomlar\u0131n\u0131n stabilitesini kontrol etmek i\u00e7in bunlar\u0131n bir oktet olu\u015fturup olu\u015fturmad\u0131\u011f\u0131n\u0131 kontrol etmemiz gerekir.<\/p>\n

Ne yaz\u0131k ki burada nitrojen atomlar\u0131 bir oktet olu\u015fturmuyor. Azotun sadece 6 elektronu vard\u0131r ve karars\u0131zd\u0131rlar. <\/p>\n

\"N2O4<\/figure>\n

\u015eimdi, bu nitrojen atomlar\u0131n\u0131 kararl\u0131 hale getirmek i\u00e7in, d\u0131\u015ftaki oksijen atomlar\u0131n\u0131n elektron \u00e7iftlerini, nitrojen atomlar\u0131n\u0131n 8 elektrona (yani bir oktete) sahip olabilece\u011fi \u015fekilde kayd\u0131rman\u0131z gerekir. <\/p>\n

\"N2O4<\/figure>\n

Bu elektron \u00e7iftlerini hareket ettirdikten sonra merkezi nitrojen atomlar\u0131 fazladan 2 elektron alacak ve b\u00f6ylece toplam elektron say\u0131s\u0131 8 olacakt\u0131r. <\/p>\n

\"N2O4<\/figure>\n

Yukar\u0131daki resimde nitrojen atomlar\u0131n\u0131n 8 elektrona sahip olmalar\u0131 nedeniyle bir oktet olu\u015fturdu\u011funu g\u00f6rebilirsiniz.<\/p>\n

Bu nedenle N2O4’\u00fcn en kararl\u0131 Lewis yap\u0131s\u0131d\u0131r.<\/p>\n

N2O4’\u00fcn yukar\u0131daki Lewis nokta yap\u0131s\u0131nda, her bir ba\u011f elektronu \u00e7iftini (:) tek bir ba\u011f<\/a> (|) olarak da temsil edebilirsiniz. Bunu yapmak, N2O4’\u00fcn a\u015fa\u011f\u0131daki Lewis yap\u0131s\u0131na yol a\u00e7acakt\u0131r. <\/p>\n

\"N2O4'\u00fcn<\/figure>\n

Umar\u0131m yukar\u0131daki t\u00fcm ad\u0131mlar\u0131 tamamen anlam\u0131\u015fs\u0131n\u0131zd\u0131r.<\/p>\n

Daha fazla pratik yapmak ve daha iyi anlamak i\u00e7in a\u015fa\u011f\u0131da listelenen di\u011fer Lewis yap\u0131lar\u0131n\u0131 deneyebilirsiniz.<\/p>\n

Daha iyi anlamak i\u00e7in \u015fu Lewis yap\u0131lar\u0131n\u0131 deneyin (veya en az\u0131ndan g\u00f6r\u00fcn):<\/strong><\/p>\n

\n\n\n\n\n\n
Lewis yap\u0131s\u0131 COF2<\/a><\/td>\n Lewis yap\u0131s\u0131 SCl4<\/a><\/td>\n<\/tr>\n
Lewis yap\u0131s\u0131 PBr5<\/a><\/td>\n Lewis yap\u0131s\u0131 SiS2<\/a><\/td>\n<\/tr>\n
OCS Lewis Yap\u0131s\u0131<\/a><\/td>\n Br3-Lewis yap\u0131s\u0131<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/figure>\n","protected":false},"excerpt":{"rendered":"

Yukar\u0131daki g\u00f6rseli zaten g\u00f6rd\u00fcn\u00fcz de\u011fil mi? Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. N2O4’\u00fcn Lewis yap\u0131s\u0131nda, merkezde d\u00f6rt oksijen (O) atomu ile \u00e7evrelenmi\u015f iki nitrojen (N) atomu bulunur. \u0130ki oksijen atomu \u00e7ift ba\u011fl\u0131, di\u011fer iki oksijen atomu ise nitrojen atomuna tek ba\u011fl\u0131. Tek ba\u011fl\u0131 oksijen (O) atomlar\u0131nda 3 ortak \u00e7ift, \u00e7ift ba\u011fl\u0131 oksijen (O) atomlar\u0131nda ise 2 ortak … Devam\u0131n\u0131 oku<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12],"tags":[],"yoast_head":"\n5 Ad\u0131mda N2O4 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/\" \/>\n<meta property=\"og:locale\" content=\"tr_TR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"5 Ad\u0131mda N2O4 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza\" \/>\n<meta property=\"og:description\" content=\"Yukar\u0131daki g\u00f6rseli zaten g\u00f6rd\u00fcn\u00fcz de\u011fil mi? Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. N2O4’\u00fcn Lewis yap\u0131s\u0131nda, merkezde d\u00f6rt oksijen (O) atomu ile \u00e7evrelenmi\u015f iki nitrojen (N) atomu bulunur. \u0130ki oksijen atomu \u00e7ift ba\u011fl\u0131, di\u011fer iki oksijen atomu ise nitrojen atomuna tek ba\u011fl\u0131. Tek ba\u011fl\u0131 oksijen (O) atomlar\u0131nda 3 ortak \u00e7ift, \u00e7ift ba\u011fl\u0131 oksijen (O) atomlar\u0131nda ise 2 ortak ... Devam\u0131n\u0131 oku\" \/>\n<meta property=\"og:url\" content=\"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/\" \/>\n<meta property=\"og:site_name\" content=\"Chemuza\" \/>\n<meta property=\"article:published_time\" content=\"2023-07-25T20:37:39+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-n2o4-lewis.jpg\" \/>\n<meta name=\"author\" content=\"Edit\u00f6r ekibi\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Yazan:\" \/>\n\t<meta name=\"twitter:data1\" content=\"Edit\u00f6r ekibi\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tahmini okuma s\u00fcresi\" \/>\n\t<meta name=\"twitter:data2\" content=\"4 dakika\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/\"},\"author\":{\"name\":\"Edit\u00f6r ekibi\",\"@id\":\"https:\/\/chemuza.org\/tr\/#\/schema\/person\/5d92b9b82b9b43578b70f5a7cd51fa3e\"},\"headline\":\"5 ad\u0131mda n2o4 lewis yap\u0131s\u0131 (resimlerle)\",\"datePublished\":\"2023-07-25T20:37:39+00:00\",\"dateModified\":\"2023-07-25T20:37:39+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/\"},\"wordCount\":898,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/chemuza.org\/tr\/#organization\"},\"articleSection\":[\"Lewis yap\u0131s\u0131\"],\"inLanguage\":\"tr\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/\",\"url\":\"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/\",\"name\":\"5 Ad\u0131mda N2O4 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza\",\"isPartOf\":{\"@id\":\"https:\/\/chemuza.org\/tr\/#website\"},\"datePublished\":\"2023-07-25T20:37:39+00:00\",\"dateModified\":\"2023-07-25T20:37:39+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/#breadcrumb\"},\"inLanguage\":\"tr\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Ev\",\"item\":\"https:\/\/chemuza.org\/tr\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"5 ad\u0131mda n2o4 lewis yap\u0131s\u0131 (resimlerle)\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/chemuza.org\/tr\/#website\",\"url\":\"https:\/\/chemuza.org\/tr\/\",\"name\":\"Chemuza\",\"description\":\"Kimyasal ke\u015fiflere a\u00e7\u0131lan kap\u0131n\u0131z!\",\"publisher\":{\"@id\":\"https:\/\/chemuza.org\/tr\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/chemuza.org\/tr\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"tr\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/chemuza.org\/tr\/#organization\",\"name\":\"Chemuza\",\"url\":\"https:\/\/chemuza.org\/tr\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"tr\",\"@id\":\"https:\/\/chemuza.org\/tr\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/chemuza.org\/tr\/wp-content\/uploads\/2023\/10\/chemuza-logo.png\",\"contentUrl\":\"https:\/\/chemuza.org\/tr\/wp-content\/uploads\/2023\/10\/chemuza-logo.png\",\"width\":387,\"height\":70,\"caption\":\"Chemuza\"},\"image\":{\"@id\":\"https:\/\/chemuza.org\/tr\/#\/schema\/logo\/image\/\"}},{\"@type\":\"Person\",\"@id\":\"https:\/\/chemuza.org\/tr\/#\/schema\/person\/5d92b9b82b9b43578b70f5a7cd51fa3e\",\"name\":\"Edit\u00f6r ekibi\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"tr\",\"@id\":\"https:\/\/chemuza.org\/tr\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/0b4093cbb11daf811af20208005bf6b5?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/0b4093cbb11daf811af20208005bf6b5?s=96&d=mm&r=g\",\"caption\":\"Edit\u00f6r ekibi\"},\"sameAs\":[\"http:\/\/chemuza.org\/tr\"]}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"5 Ad\u0131mda N2O4 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/","og_locale":"tr_TR","og_type":"article","og_title":"5 Ad\u0131mda N2O4 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza","og_description":"Yukar\u0131daki g\u00f6rseli zaten g\u00f6rd\u00fcn\u00fcz de\u011fil mi? Yukar\u0131daki g\u00f6rseli k\u0131saca a\u00e7\u0131klayay\u0131m. N2O4’\u00fcn Lewis yap\u0131s\u0131nda, merkezde d\u00f6rt oksijen (O) atomu ile \u00e7evrelenmi\u015f iki nitrojen (N) atomu bulunur. \u0130ki oksijen atomu \u00e7ift ba\u011fl\u0131, di\u011fer iki oksijen atomu ise nitrojen atomuna tek ba\u011fl\u0131. Tek ba\u011fl\u0131 oksijen (O) atomlar\u0131nda 3 ortak \u00e7ift, \u00e7ift ba\u011fl\u0131 oksijen (O) atomlar\u0131nda ise 2 ortak ... Devam\u0131n\u0131 oku","og_url":"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/","og_site_name":"Chemuza","article_published_time":"2023-07-25T20:37:39+00:00","og_image":[{"url":"https:\/\/chemuza.org\/wp-content\/uploads\/2023\/08\/structure-n2o4-lewis.jpg"}],"author":"Edit\u00f6r ekibi","twitter_card":"summary_large_image","twitter_misc":{"Yazan:":"Edit\u00f6r ekibi","Tahmini okuma s\u00fcresi":"4 dakika"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/#article","isPartOf":{"@id":"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/"},"author":{"name":"Edit\u00f6r ekibi","@id":"https:\/\/chemuza.org\/tr\/#\/schema\/person\/5d92b9b82b9b43578b70f5a7cd51fa3e"},"headline":"5 ad\u0131mda n2o4 lewis yap\u0131s\u0131 (resimlerle)","datePublished":"2023-07-25T20:37:39+00:00","dateModified":"2023-07-25T20:37:39+00:00","mainEntityOfPage":{"@id":"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/"},"wordCount":898,"commentCount":0,"publisher":{"@id":"https:\/\/chemuza.org\/tr\/#organization"},"articleSection":["Lewis yap\u0131s\u0131"],"inLanguage":"tr","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/","url":"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/","name":"5 Ad\u0131mda N2O4 Lewis Yap\u0131s\u0131 (Resimlerle) - Chemuza","isPartOf":{"@id":"https:\/\/chemuza.org\/tr\/#website"},"datePublished":"2023-07-25T20:37:39+00:00","dateModified":"2023-07-25T20:37:39+00:00","breadcrumb":{"@id":"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/#breadcrumb"},"inLanguage":"tr","potentialAction":[{"@type":"ReadAction","target":["https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/chemuza.org\/tr\/lewis-n2o4un-yapisi\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Ev","item":"https:\/\/chemuza.org\/tr\/"},{"@type":"ListItem","position":2,"name":"5 ad\u0131mda n2o4 lewis yap\u0131s\u0131 (resimlerle)"}]},{"@type":"WebSite","@id":"https:\/\/chemuza.org\/tr\/#website","url":"https:\/\/chemuza.org\/tr\/","name":"Chemuza","description":"Kimyasal ke\u015fiflere a\u00e7\u0131lan kap\u0131n\u0131z!","publisher":{"@id":"https:\/\/chemuza.org\/tr\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/chemuza.org\/tr\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"tr"},{"@type":"Organization","@id":"https:\/\/chemuza.org\/tr\/#organization","name":"Chemuza","url":"https:\/\/chemuza.org\/tr\/","logo":{"@type":"ImageObject","inLanguage":"tr","@id":"https:\/\/chemuza.org\/tr\/#\/schema\/logo\/image\/","url":"https:\/\/chemuza.org\/tr\/wp-content\/uploads\/2023\/10\/chemuza-logo.png","contentUrl":"https:\/\/chemuza.org\/tr\/wp-content\/uploads\/2023\/10\/chemuza-logo.png","width":387,"height":70,"caption":"Chemuza"},"image":{"@id":"https:\/\/chemuza.org\/tr\/#\/schema\/logo\/image\/"}},{"@type":"Person","@id":"https:\/\/chemuza.org\/tr\/#\/schema\/person\/5d92b9b82b9b43578b70f5a7cd51fa3e","name":"Edit\u00f6r ekibi","image":{"@type":"ImageObject","inLanguage":"tr","@id":"https:\/\/chemuza.org\/tr\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/0b4093cbb11daf811af20208005bf6b5?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/0b4093cbb11daf811af20208005bf6b5?s=96&d=mm&r=g","caption":"Edit\u00f6r ekibi"},"sameAs":["http:\/\/chemuza.org\/tr"]}]}},"yoast_meta":{"yoast_wpseo_title":"","yoast_wpseo_metadesc":"","yoast_wpseo_canonical":""},"_links":{"self":[{"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/posts\/148"}],"collection":[{"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/comments?post=148"}],"version-history":[{"count":0,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/posts\/148\/revisions"}],"wp:attachment":[{"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/media?parent=148"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/categories?post=148"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/chemuza.org\/tr\/wp-json\/wp\/v2\/tags?post=148"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}